Gambling: Market % Difference, Odds & Probability Explained

  • Thread starter Thread starter Yang999
  • Start date Start date
  • Tags Tags
    gambling
Yang999
Messages
1
Reaction score
0
firstly, what is the practical difference to a book maker between a market % that is less than 100% and one that is larger than 100% ?

secondly:
if i have the following prices on teams A,B,C and D:

A 1.70
B 3.00
C 8.50
D. 16.00

then are the gambling odds expressed as :
A -> "7 to 10"
B -> "2 to 7"
C -> "15 to 2"
D -> "15 to 1"

And subjective probability that team "B" wins is about 30.3%?
??

Thnx.
 
Physics news on Phys.org
Hello, I don't know what the first part of your question means, but I will answer the second.

You seem to be asking about the decimalisation of gambling odds. so let's begin by seeing what a bookmaker means by his odds. Say we bet on team A winning against team B, let us say the bookmaker gives us odds of 5/2 for a win. i.e he expresses his bet as "Team A to win 5/2" this means that for every $2 you fork out if the outcome is a success(team A wins) the bookmaker will give you 5.

So say you bet $300, that's 150 lots of $2, hence you win 150*5 $750. of course originally you also gave your man the $300 to hold onto, since he wouldn't trust you to honour your agreement and pay up,so in addition to your winnings, he has to give you your original cash back.

so total payout = (sum originally given to bookmaker)*(1+ x/y)

where x is how much you win for every y, the 1 term is there to add your original input.

i.e for your teams all your decimal odds are correct except for Team B.

3.00 odds are worth 2/1 (you fork out $5, get back $15 - which is your original 5 plus your winnings)


--onto the subjective probability, I'm not quite sure what you mean exactly. Assuming the bookmaker is offering odds on probabilities of teams winning (i.e he will break even at infinity) then Team A has a 33.333...% chance of winning.
 
Sorry I meant team B at the end there.
 
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.
Back
Top